A Level Paper 4: Mechanics, Forces and Moments
Covers proof and algebraic methods, differentiation, integration, kinematics, forces and Newton's laws, and moments.
Questions
Question 1 [2 marks]
Forces and Newton's Laws
A resultant force of 36 N gives a body an acceleration of 4 m/s^2.
Find the mass of the body.
Question 2 [2 marks]
Kinematics
A skateboarder decelerates uniformly from 8 m/s to 2 m/s with a deceleration of 1.5 m/s^2.
Find the time taken.
Question 3 [3 marks]
Differentiation
Find the coordinates of the stationary point on the curve y = x^2 - 6x + 5, and state whether it is a minimum or a maximum.
Question 4 [4 marks]
Proof and Algebraic Methods
The functions f and g are defined for all real x by f(x) = 2x - 1 and g(x) = x^2 + 3.
Find fg(2) and gf(2).
Question 5 [4 marks]
Integration
Without using a calculator, find the integral of (8x^3 - 6x + 5/x^2) with respect to x, writing 5/x^2 as 5x^-2 before integrating.
Question 6 [4 marks]
Moments
A uniform rod AB has length 7 m and is pivoted at its centre. A force of 40 N acts vertically downwards at a point 1.8 m from the pivot on one side.
Find the perpendicular distance from the pivot at which a force of 24 N must act vertically downwards on the other side for the rod to balance, and find the total moment about the pivot when the rod is balanced.
Question 7 [4 marks]
Proof and Algebraic Methods
Without using a calculator, find the quotient and remainder when 2x^3 - 3x^2 + 4x - 1 is divided by (x - 1).
Question 8 [4 marks]
Integration
The curve y = x^2 - 4x meets the x-axis at the origin and at the point (4, 0).
Find the area enclosed between the curve and the x-axis for 0 <= x <= 4.
Question 9 [3 marks]
Proof and Algebraic Methods
The function f is defined by f(x) = (2x - 1)/(x + 3) for x != -3.
Find f^-1(x) and state its domain.
Question 10 [4 marks]
Integration
A curve passes through the point (0, 5) and satisfies dy/dx = 6e^(2x) - 4.
Find the equation of the curve.
Question 11 [5 marks]
Kinematics
A stone is thrown horizontally with speed 15 m/s from the top of a cliff 20 m high. Using g = 9.8 m/s^2,
find the time taken for the stone to reach the ground, and find the horizontal distance it travels before landing.
Question 12 [4 marks]
Proof and Algebraic Methods
Without using a calculator, prove by contradiction that sqrt(3) is an irrational number.
Question 13 [5 marks]
Differentiation
Without using a calculator, differentiate y = (5x^2 - 3x)^4 with respect to x, using the chain rule.
Question 14 [3 marks]
Proof and Algebraic Methods
By finding a suitable counter-example, show that the statement "n^2 + n + 1 is a prime number for every positive integer n" is false.
Question 15 [5 marks]
Differentiation
A curve has parametric equations x = t^2, y = t^3 - 3t.
Find dy/dx in terms of t, and find the gradient of the curve at the point where t = 2.
Question 16 [6 marks]
Forces and Newton's Laws
A particle A of mass 4 kg lies on a rough plane inclined at 20 degrees to the horizontal. A is connected by a light inextensible string, passing over a smooth pulley at the top of the plane, to a particle B of mass 6 kg which hangs freely. The coefficient of friction between A and the plane is 0.25. The system is released from rest with B moving downwards.
Find the acceleration of the system and the tension in the string. Use g = 9.8 m/s^2.
Question 17 [6 marks]
Moments
A uniform ladder AB of length 6 m and weight 180 N rests with end A on rough horizontal ground and end B against a smooth vertical wall, making an angle of 60 degrees with the ground. A person of weight 700 N stands on the ladder at a point 4 m from A.
Given that the ladder is on the point of slipping, find the coefficient of friction between the ladder and the ground.
Question 18 [6 marks]
Integration
The curve y = x^3 - 4x crosses the x-axis at x = -2, x = 0 and x = 2.
Find the total area enclosed between the curve and the x-axis for -2 <= x <= 2.
Question 19 [6 marks]
Differentiation
A spherical balloon is being inflated so that its volume V cm^3 increases at a constant rate of 50 cm^3/s.
Find the rate of increase of the radius r cm when r = 5 cm. (Volume of a sphere: V = (4/3)pi r^3.)
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| using F = ma | M1 |
| m = 9 kg | A1 |
| Final answer: 9 kg | |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| using a = (v - u)/t with a = -1.5 | M1 |
| t = 4 s | A1 |
| Final answer: 4 s | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating and setting 2x - 6 = 0 | M1 |
| the point (3, -4) | A1 |
| stating it is a minimum, since d^2y/dx^2 = 2 > 0 | A1 |
| Final answer: (3, -4), a minimum point | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding g(2) = 7 | M1 |
| fg(2) = f(7) = 13 | A1 |
| finding f(2) = 3 | M1 |
| gf(2) = g(3) = 12 | A1 |
| Final answer: fg(2) = 13, gf(2) = 12 | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| rewriting 5/x^2 as 5x^-2 | M1 |
| integrating each term | M1 |
| the 2x^4 - 3x^2 terms | A1 |
| 2x^4 - 3x^2 - 5/x + c | A1 |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| taking moments about the pivot | M1 |
| forming the equation 40 x 1.8 = 24 x d | M1 |
| d = 3 m | A1 |
| the total moment = 0, since the rod is in equilibrium (the two moments are equal and opposite) | A1 |
| Final answer: d = 3 m; total moment about the pivot = 0 (equilibrium) | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| dividing to obtain the first term of the quotient, 2x^2 | M1 |
| continuing the division to find the remaining terms of the quotient | M1 |
| the quotient 2x^2 - x + 3 | A1 |
| the remainder 2 | A1 |
| Final answer: Quotient = 2x^2 - x + 3, remainder = 2 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| the antiderivative x^3/3 - 2x^2 | M1 |
| substituting the limits x = 4 and x = 0 to get -32/3 | M1 |
| recognising the integral is negative because the curve lies below the x-axis on this interval | A1 |
| the area = 32/3 square units (taking the magnitude) | A1 |
| Final answer: Area = 32/3 square units | |
| Question 9[3 marks] | |
|---|---|
| Answer or working | Marks |
| setting y = (2x - 1)/(x + 3) and rearranging to y(x + 3) = 2x - 1 | M1 |
| collecting x terms to give x(y - 2) = -1 - 3y | M1 |
| f^-1(x) = (1 + 3x)/(2 - x), with domain x != 2 | A1 |
| Final answer: f^-1(x) = (1 + 3x)/(2 - x), x != 2 | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| integrating dy/dx = 6e^(2x) - 4 to give y = 3e^(2x) - 4x + c | M1 |
| substituting x = 0 and y = 5 into the equation | M1 |
| the equation 3 + c = 5 | A1 |
| the equation of the curve y = 3e^(2x) - 4x + 2 | A1 |
| Final answer: y = 3e^(2x) - 4x + 2 | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| using s = 0.5 g t^2 vertically with s = 20 | M1 |
| t = 2.02 s (3 sf) | A1 |
| using the horizontal velocity of 15 m/s, which is unaffected by gravity | M1 |
| horizontal distance = 15 x 2.02 | M1 |
| horizontal distance = 30.3 m (3 sf) | A1 |
| Final answer: t = 2.02 s (3 sf); horizontal distance = 30.3 m (3 sf) | |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| assuming, for contradiction, that sqrt(3) is rational, so sqrt(3) = p/q where p and q are integers with no common factor | B1 |
| squaring to give p^2 = 3q^2 and deducing that p must be a multiple of 3 | M1 |
| substituting p = 3k to give q^2 = 3k^2 and deducing that q must also be a multiple of 3 | M1 |
| identifying that this contradicts p/q being in its lowest terms, so sqrt(3) is irrational | A1 |
| Final answer: Proof: assuming sqrt(3) = p/q in lowest terms forces both p and q to be multiples of 3, a contradiction, so sqrt(3) is irrational. | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the inner function u = 5x^2 - 3x and outer function u^4 | M1 |
| differentiating the outer function: 4u^3 | M1 |
| differentiating the inner function: du/dx = 10x - 3 | A1 |
| applying the chain rule dy/dx = 4u^3 x du/dx | M1 |
| dy/dx = 4(5x^2 - 3x)^3(10x - 3) | A1 |
| Final answer: dy/dx = 4(5x^2 - 3x)^3 (10x - 3) | |
| Question 14[3 marks] | |
|---|---|
| Answer or working | Marks |
| testing successive values of n (e.g. n = 1, 2, 3) and finding each gives a prime | M1 |
| testing n = 4 to obtain n^2 + n + 1 = 21 | M1 |
| identifying 21 = 3 x 7 is not prime, so the statement is false | A1 |
| Final answer: False: n = 4 gives n^2 + n + 1 = 21 = 3 x 7, which is not prime. | |
| Question 15[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding dx/dt = 2t | M1 |
| finding dy/dt = 3t^2 - 3 | M1 |
| dy/dx = (3t^2 - 3)/(2t) | A1 |
| substituting t = 2 | M1 |
| the gradient = 9/4 | A1 |
| Final answer: dy/dx = (3t^2 - 3)/(2t); gradient at t = 2 is 9/4 | |
| Question 16[6 marks] | |
|---|---|
| Answer or working | Marks |
| the equation of motion for B: 6g - T = 6a | M1 |
| resolving perpendicular to the plane for A and finding the friction force F = mu(mg cos(20)) | M1 |
| F = 9.21 N (3 sf) | A1 |
| the equation of motion for A along the plane: T - mg sin(20) - F = 4a | M1 |
| combining the two equations of motion to eliminate T | M1 |
| a = 3.62 m/s^2 (3 sf), with T = 37.1 N (3 sf) found by substitution | A1 |
| Final answer: a = 3.62 m/s^2 (3 sf); T = 37.1 N (3 sf) | |
| Question 17[6 marks] | |
|---|---|
| Answer or working | Marks |
| vertical equilibrium: R = 180 + 700 = 880 N | M1 |
| horizontal equilibrium F = S | M1 |
| taking moments about A: S x (6 sin(60)) = 180 x (3 cos(60)) + 700 x (4 cos(60)) | M1 |
| S = 321 N (3 sf) | A1 |
| using F = S at limiting equilibrium with mu = F/R | M1 |
| mu = 0.365 (3 sf) | A1 |
| Question 18[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the curve is above the x-axis for -2 < x < 0 and below it for 0 < x < 2, so the integral must be split at x = 0 | M1 |
| finding the antiderivative x^4/4 - 2x^2 | M1 |
| the integral from -2 to 0 equal to 4 | A1 |
| the integral from 0 to 2 equal to -4 | A1 |
| taking the magnitude of the second integral and adding it to the first | M1 |
| the total area = 8 square units | A1 |
| Final answer: Total area = 8 square units | |
| Question 19[6 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating V = (4/3)pi r^3 to get dV/dr = 4pi r^2 | M1 |
| using the chain rule dr/dt = dV/dt divided by dV/dr | M1 |
| substituting dV/dt = 50 | A1 |
| substituting r = 5 to get dV/dr = 100pi | M1 |
| dr/dt = 50/(100pi) = 1/(2pi) | A1 |
| dr/dt = 0.159 cm/s (3 sf) | A1 |
| Final answer: dr/dt = 1/(2*pi) = 0.159 cm/s (3 sf) | |